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2 years ago

A small hole in the wing of a space shuttle requires a 20.7-cm² patch.

Chemistry

1 answer:

cricket20 [7]2 years ago

8 0

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or, 1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 × 10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

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Pt 2. Chem Reactions 50 PTS

Dahasolnce [82]

The red colour is the limiting reactant.

Red-blue colour ball and two white balls attached together are reactants.

Red-blue colour ball and two white and one red colour ball attached to each other are products.

<h3>What is a limiting reagent?</h3>

The reactant that is entirely used up in a reaction is called a limiting reagent.

A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products.

A product is a substance that is present at the end of a chemical reaction.

Hence,

The red colour is the limiting reactant.

Red-blue colour ball and two white balls attached together are reactants.

Red-blue colour ball and two white and one red colour ball attached to each other are products.

Learn more about limiting reagents here:

brainly.com/question/26905271

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6 0

1 year ago

How did music change during medieval, renaissance and baroque eras

Vilka [71]

Answer:

usage of trills and vibrato

Explanation:

earlier music used trills while later styles would use vibrato.

7 0

3 years ago

The energy diagram illustrated in the model corresponds to what student B

KatRina [158]

Answer: Exothermic reaction

Explanation:

Exothermic reactions lose energy, or ΔH, when they react.

4 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago

Determine the molar masses for each of the following elements, compounds, or molecules. Remember to round to the proper amount o

ra1l [238]

Answer:

Explanation:

Cu(OH)2 _97.65 g/mol

Na2WO4 _293.82 g/mol

Pd _106.42 g/mol

KMnO4 _158.03 g/mol

NH3 _17.03 g/mol

3 0

2 years ago

Read 2 more answers